+% We seem not to need this lemma for multigraphs...
+%If there are any loops or parallel edges in~$G$, we can flatten the graph. According to the
+%Flattening lemma (\ref{flattening}), the MST stays the same and if we remove a parallel edge
+%or loop~$f$, then $\pi(f)$ would be removed when flattening~$G/e$, so $f$ never participates
+%in a MST.
+The right-hand side of the equality is a spanning tree of~$G$, let us denote it by~$T$ and
+the MST of $G/e$ by~$T'$. If $T$ were not minimum, there would exist a $T$-light edge~$f$ in~$G$
+(by Theorem \ref{mstthm}). If the path $T[f]$ covered by~$f$ does not contain~$e$,
+then $\pi[T[f]]$ is a path covered by~$\pi(f)$ in~$T'$. Otherwise $\pi(T[f]-e)$ is such a path.
+In both cases, $f$ is $T'$-light, which contradicts the minimality of~$T'$. (We do not have
+a~multigraph version of the theorem, but the side we need is a~straightforward edge exchange,
+which obviously works in multigraphs as well.)
+\qed
+
+\rem
+In the previous algorithm, the role of the mapping~$\pi^{-1}$ is of course played by the edge labels~$\ell$.
+
+Finally, we will show a family of graphs where the $\O(m\log n)$ bound on time complexity
+is tight. The graphs do not have unique weights, but they are constructed in a way that
+the algorithm never compares two edges with the same weight. Therefore, when two such
+graphs are monotonely isomorphic (see~\ref{mstiso}), the algorithm processes them in the same way.
+
+\defn
+A~\df{distractor of order~$k$,} denoted by~$D_k$, is a path on $n=2^k$~vertices $v_1,\ldots,v_n$
+where each edge $v_iv_{i+1}$ has its weight equal to the number of trailing zeroes in the binary
+representation of the number~$i$. The vertex $v_1$ is called a~\df{base} of the distractor.
+
+\rem
+Alternatively, we can use a recursive definition: $D_0$ is a single vertex, $D_{k+1}$ consists
+of two disjoint copies of~$D_k$ joined by an edge of weight~$k$.
+
+\figure{distractor.eps}{\epsfxsize}{A~distractor $D_3$ and its evolution (bold edges are contracted)}
+
+\lemma
+A~single iteration of the contractive algorithm reduces~$D_k$ to a graph isomorphic with~$D_{k-1}$.
+
+\proof
+Each vertex~$v$ of~$D_k$ is incident with a single edge of weight~1. The algorithm therefore
+selects all weight~1 edges and contracts them. This produces a graph which is
+exactly $D_{k-1}$ with all weights increased by~1, which does not change the relative order of edges.